Muller-Breslau’s Principle, Two hinged and Fixed Arches and Stiffening Girders

Noida Institute of Engineering and Technology,
Greater Noida
Muller-Breslau’s Principle,
Two hinged and Fixed Arches ,
and
Stiffening Girders
Aayushi
Assistant Professor
Civil Engg. Department
6/5/2022
1
Unit: 2
Aayushi RCE-502, DOS 1 Unit 2
Subject Name : Design of
Structure I
Course Details : B Tech 5th
Sem

6/5/2022 Aayushi RCE-502, DOS 1 Unit 2 2
Content
 Course Objective
 Course Outcome
 CO-PO & PSO Mapping
 Prerequisite & Recap
 Muller Breslau Principle
 Procedure for Muller Breslau principle
 Numerical & Deflected shapes
 Arches
 Types of arches
 Two hinged arches
 Numerical on two hinged arches
 Syllabus of unit 2
 Topic outcome and mapping with PO

Content
 You tube Video Links
 Daily Quiz
 Weekly Assignment
 MCQs
 Old Question Papers
 Expected Questions in University Examination
 Summary
 References
 Stiffening girder
 Fixed Arches & numerical
 Introduction &Numerical

Objective
1
To impart the principles of elastic structural analysis and behavior
of indeterminate structures.
2
To impart knowledge about various methods involved in the
analysis of indeterminate structures...
3
To apply these methods for analyzing the indeterminate structures
to evaluate the response of structures .
4
To enable the student get a feeling of how real-life structures
behave
5
To make the student familiar with latest computational techniques
and software used for structural analysis. .
6/5/2022
4
Course Objective

Students will be able
CO1 To Identify and analyze the moment distribution in beams and frames by Slope
Deflection Method, Moment Distribution Method and Strain Energy Method.
CO2 To provide adequate learning of indeterminate structures with Muller’s
Principle; Apply & Analyze the concept of influence lines for deciding the
critical forces and sections while designing..
CO3 To learn about suspension bridge, two and three hinged stiffening girders and
their influence line diagram external loading and analyze the same.
CO4 To Identify and analyze forces and displacement matrix for various structural.
CO5 To understand the collapse load in the building and plastic moment formed.
CO6 Apply the concepts of forces and displacements to solve indeterminate structure.
Course Outcome

CO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12
RCE502.1 3 3 1 2 1 - - - 2 - - 3
RCE502.2 3 2 2 2 3 - - 1 2 - 2 2
RCE502.3 3 - 2 2 1 - - - 2 - 2 2
RCE502.4 3 1 2 2 1 - - - 2 - 2 2
RCE502.5 3 2 2 2 1 - - - 2 - 2 2
RCE502.6 3 2 3 2 1 - - - 2 - 2 3
AVG. 3 2 2 2 1 - - - 2 - 2 2
CO-PO and PSO Mapping
COs PSO1 PSO2 PSO3 PSO4
RCE502.1 2 1 2 3
RCE502.2 2 1 2 2
RCE502.3 2 2 2 2
RCE502.4 2 1 2 3
RCE502.5 2 1 2 3
RCE502.6 2 1 2 3
Avg. 2 1 2 3

Prerequisite and Recap
Basics of strength of material
Basics of engineering mechanics

Syllabus of Unit 2
Muller-Breslau’s Principle and its applications
for drawing influence lines for indeterminate
beams, Analysis of two hinged and fixed arches,
Influence line diagrams for maximum bending
moment, Shear force and thrust in two hinge
arches. Analysis of two and three hinged
stiffening girders.

6/5/2022 9
Objective of Unit 2
The Muller-Breslau principle for influence lines for BM
and Shear force in beams and truss.Derivation of the
principle for different types of internal forces.

 The Müller-Breslau principle for influence lines.
 Derivation of the principle for different types of internal
forces.
 Example of application of this principle.
Topic Objective

 To Identify the moment distribution in beams and Draw ILD for
different sections.
6/5/2022 11
Topic Outcomes
Once the student has successfully completed this unit, he/she will be able
to:
 To analyze different beams and frames.
 Apply the concepts of forces and displacements to solve indeterminate
structure.

6/5/2022 12
Objective of Topic
Topic-1 Name
The Muller-Breslau principle
Objective of Topic-1:
To Construct influence line diagram (ILD) for BM and
SF in beams and truss.

6/5/2022 13
Topic Outcome and mapping with PO
Programme Outcomes (POs)
PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12
TO1 1 1 1 1 1 1 1 1 1
Outcome of Topic-1:
After the successfully competition of this topic students
will be able to Construct influence line diagram (ILD)
for BM and SF in beams and truss.

Basic concept of Influence line diagram
Influence line Diagram.
Understanding of bending moment and
shear force .
Understanding of forces.

 Heinrich Franz Bernhard Müller was born in
Wroclaw (Breslau) on 13 May1851.
 In 1875 he opened a civil engineer‘s office in Berlin.
Around this time he decided to add the name of his
hometown to his surname, becoming known as Muller-
Breslau.
 In 1883 Muller-Breslau became a lecturer and in 1885 a
professor in civil engineering at the Technische
Hochschule in Hanover.
 In 1886, Heinrich Müller-Breslau develop a method for
rapidly constructing the shape of an influence line.
Topic: History

Topic: Muller-Breslau principle
The influence line for a function (reaction, shear, moment) is to the
same scale as the deflected shape of the beam when the beam is acted
on by the function.
To draw the deflected shape properly, the ability of the beam to resist
the applied function must be removed.

Topic: Procedure for Muller-Breslau’s
Step-1:To draw ILD forany support removethat support.
Step-2:Applyunit load at that support.
Step-3: Draw bending moment diagram forthat support.
Step-4:Construct conjugate beam.
Step-5:Find deflection at some specifiedintervals
(fora conjugatebeam, deflection at any point = BM at that point).
Step-6: Divide each deflection by deflection corresponding to the point of
application of unit load.
Step-7: W
eobtain the ordinates forthe influence forthat particular support.

Step-8: For other support repeat the same
procedure.
Step-9: For ILD of bending moment,
construct a static equations from the
beam and substitute values at different
interval and you will get ordinates of
ILD for BMD.
Step-10: For shear force diagram
construct static equations and
solve.
Continuous…

Consider the following simply supported beam.
Let’s try to find the shape of the influence line for the vertical reaction
at A.
Remove the ability to resist movement in the vertical direction at A by
using a guided roller
Topic: Numerical

Remove the ability to resist movement in the vertical direction at A by
using a guided roller
Continuous…

Let’s try to find the shape of the influence line for the shear at the mid-
point (point C).
Remove the ability to resist shear at point C
Continuous…

Let’s try to find the shape of the influence line for the moment at the
mid-point (point C).
Remove the ability to resist moment at C by using a hinge
Continuous…

Topic: Explanation
 The Muller-Breslau principle uses Betti's law of virtual
work to construct influence lines. To illustrate the
method let us consider a structure AB(Figurea).
 Let us apply a unit downward force at a distance x from A ,
at point C .
 Let us assume that it creates the vertical reactions RA and RB at
supports A and B , respectively (Figure b). Let us call this
condition “System 1.”
 In “System 2” (figure c), we have the same structure
with a unit deflection applied in the direction of RA .
Here Δ is the deflection at point C .

Continuous…

 According to Betti's law, the virtual work done by the
forces in System 1 going through the Corresponding
displacements in System 2 should be equal to the virtual
work done by the forces in System 2 going through the
corresponding displacements in System 1. For these two
systems, we can write:
 The right side of this equation is zero, because in System 2
forces can exist only at the supports, corresponding to
which the displacements in System 1 (at supports A and B )
are zero. The negative sign before Δ accounts for the fact that
it acts against the unit load in System 1.
 Solving this equation we get: RA= Δ.
Continuous…
(RA)(1) + (1)(- Δ)=0

 In other words, the reaction at support A due to a unit load
at point C is equal to the displacement at point C when the
structure is subjected to a unit displacement corresponding to
the positive direction of support reaction at A .
 Similarly, we can place the unit load at any other point and
obtain the support reaction due to that from System 2.
 Thus the deflection pattern in System 2 represents
the influence line for RA.
Continuous…

ILD FORSIMPLY SUPPORTED BEAM:
Topic: Deflected Shape

ILD FORTWO SPAN CONTINOUS BEAM:
Continuous…

ILD FOR THREE SPAN CONTINOUS BEAM:
Continuous…

 In this section, we looked at main
applications for influence lines.
 The first is the use of an influence
line to determine the influence of a
single point load.
 The second is the use of an
influence line to determine the
effect of a distributed load or
patterned distributed load.
 The last is the use of an influence
line to determine the effect of a
moving pattern of loads.
Summary of Muller- Breslau

 Compute horizontal reaction in two-hinged arch by the method
of least work
 Write strain energy stored in two-hinged arch during
deformation.
 Analyse two-hinged arch for external loading.
 Compute reactions developed in two hinged arch due to
temperature loading.
Topic Objective

 To Identify the horizontal thrust and forces in the arches.
6/5/2022 32
Topic Outcomes
to:
 To analyze different Arches.
structure.

6/5/2022 33
Objective of Topic
Topic-2 Name
Arches
To analysis the strain energy stored in the two hinged
arch during deformation.

6/5/2022 34
TO1 1 1 1 1 1 1 1 1
Outcome of Topic-1:
will be able to analysis the strain energy stored in the
two hinged arch during deformation.

Briefing about Arches
Basic of force system
Basic of bending moment and shear
force

 Arches and arched structures have a wide range of uses in bridges,
arched dams and in industrial, commercial, and recreational
buildings. They represent the primary structural components of
important and expensive structures, many of which are unique.
Current trends in architecture heavily rely on arched building
components due to their strengths and architectural appeal.
 Complex structural analysis of arches is related to the analysis of the
arches strength, stability, and vibration. This type of
multidimensional analysis aims at ensuring the proper functionality
of an arch as one of the fundamental structural elements.
6/5/2022
Aayushi RCE-502, DOS 1 Unit 2 36
Topic: Arches

Topic: Types of Arches
a) Hinge less arch
b) two-hinged arch
c) one-hinged arch
d) three-hinged arch

1. Material of the arch obeys Hooke’s law (physically linear
statement)
2. Deflections of the arches are small compared with the span of
the arch (geometrically linear statement). The cases of nonlinear
statement are specifically mentioned.
3. All constraints, which are introduced into the arched structure
are two-sided, i.e., each constraint prevents displacements in two
directions. The case of one-sided constraints is specifically
mentioned.
4. In the case of elastic supports the relationship between
deflection of constraint and corresponding reaction is linear.
5. The load is applied in the longitudinal plane of symmetry of
the arch. The case of out-of-plane loading is specifically
mentioned.
Topic: Assumptions

Topic: Shape of the Arches
• Circular arch :
Ordinate y of any point of the central line of the circular arch is
calculated by the formula
Where:-
x is the abscissa of the same point of the central line of the arch;
R is the radius of curvature of the arch;
f and l are the rise and span of the arch.

Continuous…

• Analysis of two-hinged arch
Topic: Two hinged Arch

Continuous…
The strain energy due to bending
The total strain energy of the arch is given by,
according to the principle of least work ,where H is chosen as
the redundant reaction

Continuous…

Continuous…
Temperature effect

A semicircular two hinged arch of constant cross section is subjected to
a concentrated load as shown. Calculate reactions of the arch and draw
bending moment diagram.
Topic: Numerical

Taking moment of all forces about hinge B leads to,
Continuous…

Now, the horizontal reaction H may be calculated by the following
expression
Continuous…

Now the bending moment at any cross section of the arch when one of
the hinges is replaced by a roller support is given by,
Continuous…
Integrating the numerator in equation

Continuous…

Continuous…
The value of denominator in equation, after integration is,
Hence, the horizontal thrust at the support is,
Bending moment diagram
Bending moment M at any cross section of the arch is given by,

Continuous…

 The deflection and the moment at the center of the hinge less arch
are somewhat smaller than that of the two-hinged arch. However,
the hinge less arch has to be designed for support moment.
 A hinge less arch (fixed–fixed arch) is a statically redundant
structure having three redundant reactions. In the case of fixed–fixed
arch there are six reaction components; three at each fixed end.
 Apart from three equilibrium equations three more equations are
required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations
may be set up from the geometry deformation of the arch.
Topic : Fixed Arch

• Analysis of Symmetrical Hinge less Arch
Continuous…
The strain energy due to axial compression and bending

Continuous…
Consider bending moment and the axial force at the crown as the
redundant.

Continuous…
Temperature stresses
The moment at any cross-section of the arch

Strain energy stored in the arch
Continuous…
Solving equations , Mt and Ht and may be calculated

A semicircular fixed-fixed arch of constant cross section is subjected to
symmetrical concentrated load as shown .Determine the reactions of
the arch.
Continuous…

Since, the arch is symmetrical and the loading is also symmetrical,
Now the strain energy of the arch is given by,
6/5/2022
Aayushi RCE-502, DOS 1
Unit 2
59
Continuous…
The bending moment at any cross section is given by

6/5/2022
Unit 2
60
Continuous…

• Since the arch is symmetrical, integration need to be carried out
between limits 0 to π/2 and the result is multiplied by two.
Continuous…

Solving equations
Continuous…

 Two-hinged arch is the statically
indeterminate structure to degree one.
 Usually, the horizontal reaction is
treated as the redundant and is
evaluated by the Kharagpur method
of least work.
 Towards this end, the strain energy
stored in the two hinged arch during
deformation is given.
 The reactions developed due to
thermal loadings are discussed.
 Finally, a few numerical examples are
solved to illustrate the procedure.
Summary of Arches

 Differentiate between rigid and deformable structures.
 Define funicular structure.
 State the type stress in a cable.
 Analyse cables subjected to uniformly distributed load.
 Analyse cables subjected to concentrated loads.
Topic Objective

 To Identify Forces and BM in girders
6/5/2022 65
Topic Outcomes
to:
 To analyze different Girder
structure.
Analyze moments to joint rotations and support settlements.

6/5/2022 66
Objective of Topic
Topic-3 Name
The Stiffening Girder
To Differentiate between rigid and deformable
structures. Analysis of cables subjected to different
loads.

6/5/2022 67
TO1 1 1 1 1
Outcome of Topic-3:
will be able to ddifferentiate between rigid and
deformable structures. Analysis of cables subjected to
different loads.

Basic of forces in cables
Diagram of bending moment and
shear stress

The stiffening girder transmits the dead weight of the roadway and the
live traffic loads acting on the roadway in the transverse direction of
the bridge to the suspension points of the cables where these loads are
removed by the cables. As a result, horizontal compressive forces are
present in the stiffening girder
Topic : Stiffening Girder

The 3-hinged stiffening girder of a suspension bridge of span 120m
is subjected to two point loads of 240KN and 300KN at distance
25m and 80m from the left end.
a) Find the SF and BM for the girder at a distance of 40m from
the left end. The supporting cable has a central dip of 12m.
b) Find, also the maximum tension in the cable and draw the
BMD for the girder.
Topic : Stiffening Girder

Continuous…

1.To find support reaction:
Considering the stiffening girder as a suspension beam supporting the
given external load system,
ΣMa=0
240∗25+300∗80−VB∗120=0
VB=250KN
ΣFY=0
VA−240−300+250=0
VA=290KN
2.To find H:
Beam moment at C=BMc=290∗60−240∗35
BMc=9000KNm
Beam moment under the 240KN load=(290∗25)=7250KNm
Continuous…

Beam moment under the 300KN load= (250∗40)=1000KNm
Horizontal reaction at each end of the cable,H=Mc/h
H=9000/12=750KN
3.Find equivalent UDL:
Continuous…

Let we/unit= UDL transferred to the cable.
H=We∗l2/8∗h=We∗1202/8∗12=750
We=5KN/m
Each vertical reaction for the cable= V=We∗l2=5∗1202=300KN
Max tension in the cable=Tmax=(V2+H2)1/2
(3002+7502)1/2
Tmax=807.8KN
Continuous…

4. S.F calculation :
For the girder, S.F at any section = SFx=(Beamshear−Htanθ)
For the cable at any point,tanθ=4h/l2∗(l−2x)
At 40m from the left end,tanθ=4∗12/1202∗(120−2∗40)=21/5=0.1333
Beam shear at 40m from left end =(290−240)=50KN
Actual SF at 40m from the left end =(50−750∗2/15)=−50KN
5.BM calculation:
For the girder, BM at any section,
M=(Beammoment−Hmoment)=Beammoment–Hy
Beam moment at 40m from the left end =(290∗40−240∗15)=8000KNm
Continuous…

At 40m from the left end, for the cable,
y=4h/l2∗x(l−x)=4∗12/1202∗40∗80=32/3m
Actual BM at 40m from left end
=(Beammoment−Hmoment)
=(8000−750∗323)=0
6.Actual BMD for the girder:
Dip of the cable 25m from left end
=4h/l2∗x(l−x)= frac4∗121202∗25∗95=7.92m
Actual BM at 25m from left end
=(Beammoment−Hmoment)=(7250−750∗7.92)
=1310KNm
6/5/2022
Unit 2
76
Continuous…

Dip of the cable 80m from the left end =4∗12/1202∗80∗40=32/3m
Actual BM at 80m from left end =(1000−750∗32/3)=2000KNm
7.Actual BMD for the stiffening girder
Continuous…

 The girder is defined as the
structure in pure tension having
the funicular shape of the load.
 The procedures to analyse cables
carrying concentrated load and
uniformly distributed loads are
developed.
 A few numerical examples are
solved to show the application of
these methods to actual problems.
Summary of Stiffening Girder

Youtube/other Video Links
Youtube Video Links
Topic Links Process
Muller Breslau
Principle
https://www.youtube.com/watch?v
=jAuT2qaIszw&t=732s
Click on
the link
Arches https://www.youtube.com/watch?v
=jAuT2qaIszw&t=732s
Click on
the link
Stiffening Girder https://www.youtube.com/watch?v
=cPgKkWK7q28
Click on
the link

 Top most part of an arch is called ________
a) Sofit
b) Crown
c) Center
d) Abutment
 Which of the following is true in case of stone brick?
a) They are weak in compression and tension
b) They are good in compression and tension
c) They are weak in compression and good in tension
d) They are good in compression but weak in tension
Daily Quiz

 Explain Müller Breslau principle. Draw the influence line diagram
for reaction R, for the beam shown in the Fig. Compute the ordinate
at 1 m interval. The flexural rigidity is constant throughout.
 Derive the influence line diagram for reactions and bending moment
at any section of a simply supported beam. Using the ILD,
determine the support reactions and find bending moment at 2 m, 4
m and 6 m for a simply supported beam of span 8 m subjected to
three point loads of 10 kN, 15 kN and 5 kN placed at 1 m, 4.5 m and
6.5 m respectively.
Daily Quiz

 Prove that horizontal thrust developed due to a point load W acting
at crown in a two hinged semicircular arch of radius R' is
independent of its radius. Consider EI as constant.
 A two hinged parabolic arch of span 30 m and rise 6 m carries two
point loads, each 60 kN, acting at 22.5 m and 15 m from the right
end respectively. Determine the horizontal thrust and maximum
positive and negative moment in the arch.
 A parabolic two hinged arch has a span of 32 meters and a rise of 8
m. A uniformly distributed load of 1 kN/m covers 8.0 m horizontal
length of the left side of the arch. If I = I, sec 0, where 0 is the
inclination of the arch of the section to horizontal, and, I, is the
moment of the inertia of the section at the crown, find out the
horizontal thrust at hinges and bending moment at 8.0 m from the
left hinge.
Daily Quiz

 What is Indeterminate structure?
 What is a Procedure for ILD?
 What is difference between Two hinge and Fixed arches?
 Define stiffening girder.
 List out the assumptions made for Arches.
 Define Fixed arches and find out its degree of freedom.
 Define types of arches.
 Mention the formula for circular Arch.
 Define Muller- Breslau Principle.
 What is an influence line diagram? Explain its importance in
structural analysis.
Weekly Assignment

 Draw the schematic diagrams for horizontal thrust, bending moment
at any section, radial shear and normal thrust at any given section
for a typical two-hinged symmetrical parabolic arch.
 A two hinged parabolic arch has a span of 30 m and a central rise of
5.0 m . Calculate the maximum positive and negative bending
moment at a section distant 10 m from the left support , due to a
single point load of 10 kN rolling from left to right.
Weekly Assignment

 Shape of three hinged arch is always :-
a) Hyperbolic
b) Circular
c) Parabolic
d) Can be any arbitrary curve
 Internal bending moment generated in a three hinged arch is
always:-
a) 0
b) Infinite
c) Varies
d) Non zero value but remains constant
MCQ s

 What is the degree of indeterminacy of a fixed arch?
a) 1
b) 2
c) 3
d) 4
 What is the degree of indeterminacy of a two hinged arch?
a) 1
b) 2
c) 3
d) 4
MCQ s

 Which of the following is true in case of stone brick?
a) They are weak in compression and tension
b) They are good in compression and tension
c) They are weak in compression and good in tension
d) They are good in compression but weak in tension
MCQ s

 Identify the FALSE statement from the following, pertaining to
the methods of structural analysis.
1. Influence lines for stress resultants in beams can be drawn using
Muller Breslau's Principle.
2. The Moment Distribution Method is a force method of analysis, not
a displacement method.
3. The Principle of Virtual Displacements can be used to establish a
condition of equilibrium.
4. The Substitute Frame Method is not applicable to frames subjects
to significant side sway.
MCQ s

Old Question Papers

 The 3-hinged stiffening girder of a suspension bridge of span 120m
is subjected to two point loads of 240KN and 300KN at distance
25m and 80m from the left end. Find the SF and BM for the girder
at a distance of 40m from the left end. The supporting cable has a
central dip of 12m. Find, also the maximum tension in the cable and
draw the BMD for the girder.
 A semicircular two hinged arch of constant cross section is subjected
to a concentrated load as shown. Calculate reactions of the arch and
draw bending moment diagram.
Expected Questions for University Exam

 Using Muller Breslau Principle, compute the influence line
ordinates at 2 m intervals for moment at mid span of BC of the
continuous beam ABC shown in Fig.
 A two hinged semicircular arch of radius R' carried l a load "W' at a
section the radius vector corresponding to which makes an angle 'a'
with the horizontal. Find the horizontal thrust at each support.
Assume uniform flexural rigidity.
Expected Questions for University Exam

 Introduction- Muller Breslau principle , Arches i.e. Two
hinged, fixed hinged arches and Stiffening Girder.
 Explanation of procedure for all methods.
 Illustrate some examples to understand concept better.
Summary

 Jain, A. K., “Advanced Structural Analysis “, Nem Chand
& Bros., Roorkee.
 Hibbeler, R.C., “Structural Analysis”, Pearson Prentice
Hall, Sector - 62, Noida-201309
 C. S. Reddy “Structural Analysis”, Tata Mc Graw Hill
Publishing Company Limited,New Delhi.
 Timoshenko, S. P. and D. Young, “ Theory of Structures”
, Tata Mc-Graw Hill BookPublishing Company Ltd., New
Delhi.
 Dayaratnam, P. “ Analysis of Statically Indeterminate
Structures”, Affiliated East-WestPress.
 Wang, C. K. “ Intermediate Structural Analysis”, Mc
Graw-Hill Book PublishingCompany Ltd.
References

References

Muller-Breslau’s Principle, Two hinged and Fixed Arches and Stiffening Girders

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